Respuesta :
Given:
[tex]h(t)=-16t^2+96t+6[/tex]Find-:
(a) Maximum second after launch will the object reach its maximum height
(b) Find the maximum height that the object reaches.
(c) Find the x-intercept and explain its meaning in the context of the problem.
(d) After how many seconds will the object be 100 feet above the ground
(e) Find the y-intercept and explain its meaning on the context of the problem
Sol:
(a)
Maximum second after launch.
For maximum value derivative should be zero.
[tex]\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h^{\prime}(t)=-(16\times2)t+96 \\ \\ \end{gathered}[/tex][tex]\begin{gathered} -32t+96=0 \\ \\ 32t=96 \\ \\ t=\frac{96}{32} \\ \\ t=3 \end{gathered}[/tex]After 3-second the object reaches maximum height.
(b)
For maximum height is at t = 3
[tex]\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h(3)=-16(3)^2+96(3)+6 \\ \\ h(3)=(-16\times9)+(96\times3)+6 \\ \\ h(3)=-144+288+6 \\ \\ =150 \end{gathered}[/tex](c) x-intercept the value of y is zero that means:
[tex]\begin{gathered} h(t)=0 \\ \\ -16t^2+96t+6=0 \\ \\ -8t^2+48t+3=0 \\ \\ t=\frac{-48\pm\sqrt{48^2-4(-8)(3)}}{2(-8)} \\ \\ t=\frac{-48\pm48.98}{-16} \\ \\ t=6;t=-0.061 \end{gathered}[/tex]The negative value of "t" is not considered so at
x-intercept is 6 and -0.061
(d) Object be 100 feet above grounded is:
[tex]\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ -16t^2+96t+6=100 \\ \\ -16t^2+96t-94=0 \\ \\ -8t^2+48t-47=0 \\ \end{gathered}[/tex]So, the time is:
[tex]\begin{gathered} t=\frac{-48\pm\sqrt{48^2-4(-8)(-47)}}{2(-8)} \\ \\ t=\frac{-48\pm\sqrt{800}}{-16} \\ \\ t=\frac{-48-28.28}{-16},t=\frac{-48+28.28}{-16} \\ \\ t=4.76,t=1.23 \end{gathered}[/tex]At t= 4.76 and t =1.23
(e)
For y-intercept value of "x" is zero.
[tex]\begin{gathered} h(t)=-16t^2+96t+6 \\ \\ h(0)=-16(0)^2+96(0)+6 \\ \\ h(0)=6 \end{gathered}[/tex]So, the y-intercept is 6.
